The oblique shock calculator above solves the classical supersonic wedge problem. When a supersonic stream is turned into itself by a wedge, a ramp or a compression corner, it cannot turn gradually the way a subsonic flow does. It turns abruptly across a thin, inclined discontinuity, and everything about the gas changes in the space of a few mean free paths: pressure rises, temperature rises, density rises, velocity falls, and stagnation pressure is irrecoverably lost.
Arb Digital publishes free engineering calculators that name the relation they implement instead of presenting a number as if it fell from the sky. This page solves the theta–beta–Mach relation numerically for the shock angle, then applies the standard normal-shock relations to the component of velocity perpendicular to the shock. It handles both roots, tells you when a deflection is impossible, and states clearly where the perfect-gas assumptions behind it stop being trustworthy.
What This Oblique Shock Calculator Does
The central trick of oblique shock theory is that an oblique shock is a normal shock in disguise. Decompose the upstream velocity into a component normal to the shock front and a component tangential to it. The tangential component passes through unchanged. The normal component behaves exactly like a normal shock, obeying the same jump conditions. So every oblique shock property follows from the normal Mach number Mn1 = M₁ sin β.
The difficulty is finding β. The relation linking deflection θ, shock angle β and upstream Mach number cannot be inverted in closed form for β, so the calculator solves it numerically. It first locates the deflection maximum for your Mach number by search, which tells it whether a solution exists at all, then bisects on the branch you selected.
The hero result is the shock angle in degrees. The grid gives the downstream Mach number, the static pressure ratio, the static temperature ratio and the total pressure recovery. The note underneath reports the density ratio, the maximum deflection your Mach number can sustain, and the Mach angle, which is the weak-shock limit as deflection goes to zero.
How to Use It
- Enter the deflection, not the shock angle. For a symmetric wedge aligned with the flow, the deflection is the half-angle. For a wedge at incidence the two surfaces see different deflections and need two separate runs.
- Leave the branch on weak unless you have a reason. The weak solution is the one nature selects in essentially all external flow. The strong solution requires a downstream pressure high enough to force it, which is a condition you impose, not one that arises.
- Check the maximum deflection reported in the note. If your wedge angle exceeds it, no attached oblique shock exists and the shock detaches into a curved bow wave standing ahead of the body.
- Set γ for your gas and temperature range. The default 1.4 is right for air up to a few hundred kelvin. Above that, vibrational modes and eventually dissociation lower the effective value and the perfect-gas relations drift.
- Add upstream pressure and temperature if you want absolute numbers. They have no effect whatsoever on the ratios, which is one of the more elegant results in compressible flow.
The Formula and a Worked Example
The theta–beta–Mach relation, published in the standard form used by NASA Glenn Research Center's Oblique Shock Waves page, links the three angles: tan θ = 2 cot β (M₁² sin²β − 1) ÷ (M₁²(γ + cos 2β) + 2). For each M₁ the right-hand side rises from zero at the Mach angle, reaches a maximum, and falls back to zero at β = 90°. Two shock angles therefore give the same deflection, which is the origin of the weak and strong solutions.
Once β is known, the normal Mach number Mn1 = M₁ sin β drives the jump conditions. Pressure ratio is p₂/p₁ = 1 + 2γ(Mn1² − 1)/(γ + 1). Density ratio is (γ + 1)Mn1² ÷ ((γ − 1)Mn1² + 2). Temperature ratio is the pressure ratio divided by the density ratio, from the perfect gas law. The downstream normal Mach number follows from Mn2² = (1 + ½(γ−1)Mn1²) ÷ (γMn1² − ½(γ−1)), and the full downstream Mach number is M₂ = Mn2 ÷ sin(β − θ).
Work the default through. At M₁ = 3 with a 20° deflection and γ = 1.4, the weak root is β = 37.76°. Then Mn1 = 3 sin 37.76° = 1.837. The pressure ratio is 1 + (2.8/2.4)(1.837² − 1) = 1 + 1.1667 × 2.3746 = 3.770. The density ratio is 2.4 × 3.3746 ÷ (0.4 × 3.3746 + 2) = 8.099 ÷ 3.350 = 2.418, so the temperature ratio is 3.770 ÷ 2.418 = 1.559. The downstream normal Mach number is √(1.6749 ÷ 4.5244) = 0.6084, and dividing by sin(37.76° − 20°) = 0.3050 gives M₂ = 1.994. Those are the values printed in every gas dynamics table for this case.
Why an Oblique Shock Beats a Normal Shock in an Inlet
Compare the same free stream processed two ways. A normal shock at Mach 3 recovers only about 32 % of the stagnation pressure and leaves the flow at Mach 0.475. The 20° oblique shock above recovers about 80 % and leaves the flow at Mach 1.99. That difference is the entire reason supersonic inlets are built with ramps and cones instead of a plain hole.
The mechanism is straightforward once you see the decomposition. Losses across a shock depend only on the normal Mach number, and inclining the shock reduces the normal component while leaving the tangential component untouched. A shallower shock is a weaker shock. Chain several of them and you can decelerate the flow in stages, each one cheap, before finishing with a weak terminal normal shock at a Mach number close to one, where losses are almost nil.
This is why external compression inlets on supersonic aircraft carry two or three ramps. Each ramp turns the flow a little more and each shock costs a little stagnation pressure, but the product of several high recoveries beats a single low one by a wide margin. The trade is drag and complexity: every degree of turning is also a degree of flow the airframe has to un-turn later. The Mach number calculator handles the upstream free-stream conditions that set the problem up, and the isentropic flow calculator covers the loss-free portions of the duct between shocks.
Detachment: When No Attached Shock Exists
For every upstream Mach number there is a maximum deflection the flow can turn through with an attached oblique shock. At Mach 2 it is about 23°; at Mach 3 about 34°; at Mach 5 about 41°; and in the infinite-Mach limit it approaches roughly 45.6° for γ = 1.4. Ask for more than that and the shock cannot stay attached to the leading edge. It detaches and stands off as a curved bow shock ahead of the body.
A detached shock is a genuinely different flow. It is normal to the streamline at the stagnation point and weakens progressively as it curves away, so the flow behind it is subsonic near the centreline and supersonic further out, separated by a sonic line. Total pressure loss varies from point to point along the shock, the entropy layer that results is non-uniform, and none of the closed-form relations on this page apply. The calculator refuses to return numbers in that case and says so, rather than reporting a root that does not exist.
Blunt bodies live permanently in this regime by design. A re-entry capsule wants a strong detached bow shock precisely because it dumps energy into the shock layer rather than into the vehicle's surface. Sharp supersonic leading edges want the opposite. Which one is right depends entirely on whether you are optimising for drag or for heat load.
Where the Perfect-Gas Assumptions Fail
The relations implemented here assume a calorically perfect gas: constant specific heats, no vibrational excitation, no dissociation, no ionisation, no chemistry. They also assume the shock is infinitesimally thin, that the flow is steady, and that viscosity matters only inside the shock itself.
Those assumptions hold well for air up to about Mach 4 or 5 at ordinary atmospheric temperatures. Above that, the temperature behind the shock climbs high enough that vibrational modes in nitrogen and oxygen absorb energy, and the effective γ falls below 1.4. Higher still and the molecules dissociate, which absorbs enormous energy and makes the real temperature rise far smaller than the perfect-gas prediction while the density rise is far larger. Hypersonic vehicle design uses equilibrium or non-equilibrium chemistry models for exactly this reason, and a perfect-gas oblique shock calculation at Mach 12 will be wrong by a margin that matters.
Viscous interaction is the other limit. Near a surface, the shock meets the boundary layer and the interaction can separate the flow entirely, replacing your clean single shock with a lambda-shaped structure and a separation bubble. MIT OpenCourseWare's 16.100 Aerodynamics course covers the inviscid theory and the boundary-layer interactions that limit it, which is the right next step if this page's assumptions look shaky for your problem.
Arb Digital builds free tools like this one because useful pages earn attention. If you want tools, calculators or content built for your own audience, we can help.
Browse All Free Tools Talk to Arb DigitalCommon Mistakes to Avoid
- Entering the shock angle instead of the deflection — the input is the angle the flow is turned through, which is always the smaller of the two and equals the wedge half-angle for a symmetric wedge.
- Taking the strong solution by default — external supersonic flow selects the weak root unless downstream back pressure forces otherwise, and the strong root gives a subsonic answer that will not match a wind tunnel.
- Ignoring the detachment limit — past the maximum deflection there is no attached shock at all, and any number a solver returns for that case is spurious.
- Using γ = 1.4 at hypersonic speeds — vibrational excitation and dissociation lower the effective ratio of specific heats and break the perfect-gas relations well before Mach 10.
- Reading static pressure ratio as a performance figure — the number that governs inlet and engine performance is total pressure recovery, and a large static rise usually means a large total loss.
Related Free Tools From Arb Digital
Set the upstream state with the Mach number calculator and the speed of sound calculator, and get ambient conditions from the air pressure at altitude calculator and the air density calculator. For the loss-free stretches of a duct use the isentropic flow calculator, and for the opposite process — supersonic flow turning away from itself and accelerating — the Prandtl–Meyer expansion calculator is the mirror image of this page. Forces on the body itself come from the drag force calculator and the lift coefficient calculator, while the Bernoulli equation calculator covers the incompressible limit where none of this applies. The full list is on the free online tools hub.
Frequently Asked Questions
Both satisfy the theta-beta-Mach relation for the same deflection. The weak solution has a shallower shock angle, smaller losses and leaves the flow supersonic in almost every case. The strong solution has a steeper angle, larger losses and leaves the flow subsonic. External supersonic flow selects the weak root unless downstream back pressure forces the strong one.
Because there is a maximum deflection any given Mach number can turn through while keeping a shock attached to the leading edge. Beyond it no attached solution exists, so the shock moves upstream and stands off as a curved bow wave. Behind that bow shock the flow is subsonic near the centreline and supersonic further out, and the simple closed-form relations no longer apply.
Because the losses depend only on the velocity component normal to the shock front. Inclining the shock reduces that component while leaving the tangential component untouched, so a shallower shock is a weaker shock. Chaining two or three oblique shocks before a weak terminal normal shock recovers far more stagnation pressure than a single normal shock would.
It changes the absolute downstream pressure and temperature but not a single one of the ratios. Pressure ratio, density ratio, temperature ratio, total pressure recovery, shock angle and downstream Mach number all depend only on the upstream Mach number, the deflection angle and the ratio of specific heats. That is why gas dynamics tables can be printed without reference to altitude.
The Mach angle is the arcsine of one over the Mach number, and it is the angle of an infinitesimally weak disturbance. It is the lower limit of the shock angle: as the deflection tends to zero the weak shock angle tends to the Mach angle. Any real oblique shock is steeper than the Mach angle, and the difference grows with deflection.
Yes, in a narrow band of deflections just below the maximum. The deflection at which the downstream Mach number reaches one is slightly less than the deflection at which the shock detaches, so there is a small range where the weak solution still exists but leaves the flow subsonic. The calculator reports the downstream Mach number so you can see when that happens.
They are reliable for air up to roughly Mach 4 or 5 at ordinary atmospheric temperatures, where the calorically perfect gas assumption holds. Above that, vibrational excitation lowers the effective ratio of specific heats and dissociation absorbs large amounts of energy, so real temperature rises are much smaller and density rises much larger than the perfect-gas prediction.
This tool is provided for educational and preliminary design use. It applies inviscid, calorically perfect gas relations to a straight, steady, attached oblique shock, and models no real-gas chemistry, boundary layer interaction, three-dimensional effects or shock curvature. Verify any design against validated computational fluid dynamics or wind tunnel data.
